Biology 105 - Summer 2013 MIDTERM EXAM Name: SCORE: 5 X 2 Point Questions 10 points 12 X 5 Point Questions 60 points 4 X 10 Point Questions 40 points Drop the lowest of the 10 point questions -10 points TOTAL POSSIBLE SCORE 100 points
2 POINT QUESTIONS 1. [2 POINTS] How many different types (genoytypes) of gamete could be formed by an individual that is Aa BB cc Dd? (2) (1) (1) (2) = 4 2. [2 POINTS] The map distance between gene A and gene B is 10 map units. The map distance between gene B and gene C is 20 map units. However, in a testcross of an individual heterozygous for genes A and C only 26% of the progeny are recombinant. Briefly explain how can you reconcile the result of the testcross with the genetic map. A and C must be 30 map units (cm) apart. In a cross, they exhibit only 26% recombination due to the effect of multiple crossovers. 3. [2 POINTS] If you have a DNA molecule with 1,000 base pairs (1 kb) whose GC% value is 60%, how many hydrogen bonds will there be between the two strands? (600 X 3) + (400 X 2) = 2,600 4. [2 POINTS] Which of the following treatments will be likely to stabilize the DNA double helix and reduce the conversion to single strands? Check any/all that apply. raising the temperature _X_ raising the Mg ++ concentration _X_ increasing the %GC in the DNA sequence _X_ removing or inactivating helicases _X_ allowing basic proteins (histones) to bind to the DNA 5. [2 POINTS] In the Charlie Chaplain paternity case the mother was blood type A and the child was blood type B. a. Which blood type phenotypes are possible for the father? B or AB b. What is the mother s genotype? I A i c. What is the child s genotype? I B i
5 POINT QUESTIONS 6. [5 POINTS] The red vizcacha rat (Tympanoctomys barrerae) is a species of rodent in the family Octodontidae. The rodent is not a rat, but related to guinea pigs and chinchillas. This species has the highest number of chromosomes known among mammals; somatic cells have 2n = 102. How many chromatids and chromosomes are present # CHROMATIDS # CHROMOSOMES a. at anaphase of mitosis in an adult somatic cell 204 204 b. during G1 prior to meiosis I 102 102 c. in a sperm cell 51 51 d. during G2 of interphase in an adult somatic cell 204 102 e. in secondary oocyte 102 51 7. [5 POINTS] In Drosophila, the following pairs of alleles are known: Wing size: Eye Shape: t = recessive allele for tiny wings T dominant allele for normal wings n = recessive allele for narrow eyes N = dominant allele for oval (normal) eyes For the two genetic crosses below (A. and B.) determine the genotypes of each parent by analyzing the results of the cross. You should assume that the genes are NOT linked. A. PARENT 1 Phenotype PARENT 2 Phenotype OFFSPRING Phenotypes normal oval X normal oval 900 normal oval 300 normal narrow 300 tiny oval 100 tiny narrow PARENT A1 GENOTYPE: Tt Nn PARENT A2 GENOTYPE: Tt Nn B. PARENT 1 Phenotype PARENT 2 Phenotype OFFSPRING Phenotypes tiny oval X normal narrow 500 normal oval 500 tiny oval PARENT B1 GENOTYPE: tt NN PARENT B2 GENOTYPE: Tt nn
8. [5 POINTS] In wild populations of foxes there are a few individuals with silver fur rather than the typical reddish-brown fur. A fox rancher who sells pelts to the fashion industry realizes that silver pelts would fetch a higher price than the ordinary brown and so initiates a breeding program to develop a pure bred line of silver foxes. Pairs of silver foxes are trapped in the wild and mated to each other in captivity where they always produce litters that are 1/3 brown and 2/3 silver. Thereafter, when the silver foxes born in captivity are mated to each other they always produce litters that are 1/3 brown and 2/3 silver just as their parents did. Therefore the rancher fails to produce a pure breeding line of silver foxes. This is surprising because the brown foxes always breed true for brown. Another interesting feature of the mating of silver foxes was that they produced smaller litters than matings between brown foxes. Propose (briefly) a genetic explanation for this episode. The allele that causes silver fur in the heterozygous condition is a recessive lethal. This leads to the modified monohybrid ratio of 1/3 : 2/3 as follows: P: Ss [SILVER] X Ss [SILVER] F1: 1/4 SS [BROWN] 1/2 Ss [SILVER] 1/4 ss [LETHAL] Use your explanation to predict the result of a cross between a brown and silver fox. Cross: SS [BROWN] X Ss [SILVER] Predict: 1/2 SS [BROWN] 1/2 Ss [SILVER] normal litter size 9. [5 POINTS] This problem involves 2 genes, genes A and B which are 8 map units (cm) apart on the same chromosome. You begin with 2 homozygous individuals with genotypes AA bb and aa BB. (a and b are recessive alleles.) These homozygous individuals are crossed (to each other) to obtain an F1 progeny. The F1s are then testcrossed. What % of the progeny of this testcross progeny will be homozgous recessive for both genes? 4 %
10. [5 POINTS] In box A. insert the correct name for the BASE (not the name of the nucleotide). A. = Cytosine In box B. correctly indicate the direction/"polarity" of the DNA strand. B. = 5' B. A. 11. [5 POINTS] A. If you expose a culture of growing cells to radioactively labeled thymidine ( 3 H-thymidine) during S phase, how would the radioactive label be distributed over a pair of homologous chromosomes at the next mitotic metaphase? The radioactive label would appear in d (Insert the best answer from the choices below.) a. one chromatid of one homolog b. both chromatids of one homolog c. one chromatid of each homolog d. both chromatids of both homologs e. random locations B. If you remove the labeled thymidine after S phase, and then let the cells go through another S phase in the labeled thymidine how would it be distributed now at the following mitotic metaphase? The radioactive label would appear in c (Insert the best answer from the choices above.) 12. [5 POINTS] Our model for DNA replication in E. coli invoked 2 different DNA Polymerase enzymes, Pol I and Pol III. While both enzymes are capable of elongating the 3' end of a pre-existing strand, Pol I has an important function that Pol I does not have. Describe the special role of DNA Pol I. Pol I is a 5' to 3' exonuclease that removes the RNA primers used in dioscontinuous synthesis on the lagging strand (i.e. from the Okazaki pieces).
13. [5 POINTS] Matching. For each listed function choose the correct E. coli protein. Write the letter that identifies the protein in the correct box of the column provided. Each protein should be used exactly once. FUNCTION 5' 3' DNA template-dependent DNA polymerase responsible for the majority of leading and lagging strand DNA synthesis Seals (repairs) single-strand nicks remaining between adjacent Okazaki fragments in discontinuous replication on lagging strand Stabilizes DNA single-strands/prevents them from re-forming double-stranded DNA Removes ("relaxes") the positive supercoiling introduced by helix unwinding An enzyme with 3' 5' DNA polymerase activity (i.e. can add nucleotides to the 5' end of a strand). Unwinds the DNA ahead of the replication fork at rates approaching 1,000 bp/second. Requires ATP hydrolysis. An RNA polymerase that makes the primers for synthesis of Okazaki fragments 5' 3' DNA template-dependent DNA polymerase responsible for synthesizing DNA on the lagging strand only 14. [5 POINTS] Identify the protein by writing the correct letter F D G A H C B E PROTEINS A. Topoisomerase B. Primase C. Helicase D. DNA Ligase E. DNA Polymerase I F. DNA Polymerase III G. Single-Strand DNA Binding Protein H. NONE, does not exist. The diagram below shows two genes, Gene A and Gene B. The complementary DNA stands are designated Watson and Crick. Gene A is transcribed from right to left using Watson as the template strand. Gene B is transcribed from the promoter PB whose location is shown. Watson Crick Gene A P B Gene B a. Which strand (W or C) is the template for transcription of gene B? (WATSON or CRICK) b. Label the right-hand end of the CRICK DNA strand as either 5' or 3'. 5'
15. [5 POINTS] Describe (in as much detail as you can muster) the molecular structure of the 5' end of a mature (processed) eukaryotic mrna. The 5' end has methylguanine attched via its 5' carbon (i.e. "backwards") via 3 phosphates, to the 5' carbon of the first nucleotide in the transcript. See Fig. 8.13. Partial credit for "methylguanine".
10 POINT QUESTIONS 16. [10 POINTS] Assume you have 2 homozygous, true-breeding, varieties of a hypothetical organism. One variety is always RED when inbred and the other variety is always WHITE. When you cross these varieties to each other, the F1 generation consists entirely of RED individuals. When the RED F1's are inbred (crossed among themselves) you obtain a population of F2's with a phenotypic ratio of 15 RED to 1 WHITE. Work out the genetics of the inheritance of the color difference in these two varieties and predict the result of a cross between a RED F1 individual and their WHITE homozygous parent. The best prediction is that the phenotypic ratios in the progeny of this cross will be: This is a dihybrid cross in which the two genes are redundant in the sense that a single dominant allele of either gene will express the RED phenotype. WHITE is expressed only by individuals that are homozygous recessive at both genes. A_ B_ 9/16 A_ bb 3/16 aa B_ 3/16 aa bb 1/16 RED RED RED WHITE So testcrossing the heterozygous F1s gives 3/4 RED and 1/4 WHITE. 17. [10 POINTS] In this problem assume the following: There is a gene affecting height. Expression of a dominant allele (H) gives tall and expression of the recessive allele (h) gives short. There is a second gene affecting girth. Expression of a dominant allele gives thick (G) while expression of the recessive allele (g) gives thin. You locate a tall, thick individual in nature and in order to determine the genotype of this individual you perform a testcross. The results of the testcross are: 501 tall, thick 499 short, thick Next, you take a tall, thick individual from the progeny of the first testcross and use them to perform a second testcross. The results of the second testcross are: 450 tall, thick 450 short, thin 50 tall, thin 50 short, thick Explain these genetic results as fully as possible. In other words, what information can you deduce from the results of the 2 testcrosses? 1. The tall, thick individual from nature was Hh GG. 2. The genes are linked and 10 map units (cm) apart.
18. [10 POINTS] MATCHING COLUMN A COLUMN B Physical pairing of homologous chromosomes during Meiosis I. i a. epistasis Sequence recognized by an RNA polymerase as a transcription m b. Barr Body initiation site Alternate forms of a single gene h c. dominant Alleles of one gene mask or modify the expression of alleles of a d. introns a different gene Functionally inactivated X chromosome in cells of mammalian b e. testcross females Subunits (components) of spliceosomes l f. poly A tailing The allele expressed in the phenotype of a heterozygous c g. purine individual Sequences removed from an RNA transcript before translation d h. alleles Process that modifies the 3' end of eukaryotic pre-mrna f i. synapsis transcripts A nitrogenous base containing a double ring found in nucleic g j. thymidine acids A nucleotide in DNA containing a single ring in the base. j k. pleiotrpy A genetic cross in which one of the two parents is known to be e l. snrnps homozygous recessive One gene affecting more than one phenotypic character k m.promoter 19. [10 POINTS] Red-green colorblindness in humans is due to the expression of a recessive allele of an X- linked gene. A woman with normal vision whose mother was color-blind has a child fathered by a color-blind man. What is the probability that their first child will be color blind? The mother must be heterozygous (X C X c ). So 1/2 the offspring would be colorblind.
20. [10 POINTS] The pedigree diagram below shows the inheritance of a rare liver disease. A. What is the most likely genetic basis for the inheritance of this disease (check one option below)? 4 pts. autosomal recessive autosomal dominant X-linked recessive X-linked dominant X B. If individuals 1 and 2 have a child, what is the probability that their first child will have the disease? 6 pts. Since this is a RARE condition we can assume that the exogamous individuals (those that marry into the family) do not carry the allele responsible (they are homozygous for the normal dominant allele). In this case the probability that individual 1 inherits the recessive allele is 1/2. On the other hand, individual 2 must be heterozygous because one of their parents was homozygous. So the probability that a child of 1 X 2 is affected would be (1/2) (1) (1/4) = 1/8